Oxymercuration–demercuration: practice and mechanism
Oxymercuration adds water with Markovnikov orientation (OH on the more substituted carbon), but via a mercurinium ion instead of a free carbocation. That is why there are no rearrangements.
Example
Hg²⁺ forms a mercurinium ion with the alkene; water attacks the more substituted carbon (Markovnikov). NaBH₄ replaces the mercury with hydrogen. Unlike acid-catalysed hydration no free carbocation forms, so there are no rearrangements.
Mechanism step by step
- 1
Hg(OAc)₂ adds to the double bond and forms a bridged mercurinium ion (similar to the bromonium ion).
- 2
Water attacks the more substituted carbon (more partial positive charge), anti to the mercury. Deprotonation gives an organomercury alcohol.
- 3
NaBH₄ replaces the mercury by H (demercuration). The Markovnikov alcohol remains.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Background
The two steps
First mercury(II) acetate in water/THF adds to the double bond (oxymercuration); water opens the mercurinium ion at the more substituted carbon. Then sodium borohydride in basic solution replaces the C–Hg bond with C–H (demercuration).
Compared with other hydrations
Acid-catalysed hydration also gives the Markovnikov alcohol but goes through a free carbocation and can rearrange. Hydroboration–oxidation gives the anti-Markovnikov alcohol instead. Oxymercuration is the route to the Markovnikov product without rearrangement.
Variants and drawback
With an alcohol instead of water an ether forms (alkoxymercuration). The big drawback is the toxic mercury compounds, so the reaction is used less in the lab today, but it remains an exam classic.
Typical mistakes
Often confused with: Hydroboration–oxidation · Acid-catalysed hydration
Frequently asked questions
What is the advantage over H₂O/H⁺?
There is no free carbocation, so no rearrangements. The conditions are also mild.
How do I get the anti-Markovnikov alcohol?
With hydroboration–oxidation (BH₃, then H₂O₂/NaOH): the OH group then ends up on the less substituted carbon.
Why is there no rearrangement in oxymercuration?
A bridged mercurinium ion forms instead of a free carbocation; water attacks the more substituted carbon without any shift.